The infinite group of Higman which has no finite quotient is given by the presentation (with 4 generators and 4 relations): $$ G = \langle a_i, i \in \mathbb{Z}/4\mathbb{Z} \mid a_ia_{i+1 \,(\text{mod } 4)}a_i^{−1}=a^2_{i+1} ⟩. $$ Its main feature (proved by Higman) is that [it is infinite and] it has no finite quotients. Another important feature (proved by Schupp, see this post) is that it is SQ-universal (:="every countable group can be embedded in one of its quotient groups").

My question is: is it known whether this group has property (T) or not?

My lame attempts to get an insight led me to a first side question: does SQ-universal exclude property (T)?

Then I was looking into an algorithmic way of trying property (T) on Higman's group (Zuk's criterion). Which brought to the other side question: has this group a solvable word problem ? (Superficially, it looks like it could, by analogy with the Baumslag-Solitar groups).

Lastly, a presentation is rarely enough to get any information on the representation theory of the group. Are there any known facts about the representations of Higman's group (except it has no finite-dimensional non-trivial representations [using finite generation + Malcev theorem to the effect that finitely generated linear groups are residually finite])

  • 8
    $\begingroup$ No it does not have property T, since it is a nontrivial amalgam (see mathoverflow.net/questions/138791/…). $\endgroup$ – YCor Oct 16 '15 at 16:37
  • 3
    $\begingroup$ Second, non-elementary hyperbolic groups are always SQ-universal and sometimes have Property T. $\endgroup$ – YCor Oct 16 '15 at 16:38
  • 3
    $\begingroup$ Third, it probably has a solvable word problem, using this amalgam decomposition $A\ast_C B$; the point is to show that both $A$ and $B$ have solvable word problem (easy, in turn writing them as amalgam of two Baumslag-Solitar groups $BS(1,2)$ over cyclic subgroups), and to show that the amalgamated subgroup has recognizable membership problem (I haven't checked but it's probably an exercise). $\endgroup$ – YCor Oct 16 '15 at 16:41
  • $\begingroup$ @YCor : Is it known that any cocompact discrete subgroup of $\mathrm{O}_{1,n}(\mathbf R)$ is SQ universal ? I would be glad if you could give a reference ... $\endgroup$ – few_reps Oct 16 '15 at 18:45
  • 2
    $\begingroup$ @few_reps it's written in the Wikipedia page in an even broader generality (every relatively hyperbolic group, so this includes non-cocompact lattices as well). In the hyperbolic case SQ-universality is due to (independently) Delzant and Olshanskii. $\endgroup$ – YCor Oct 16 '15 at 19:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.